Level 3 — ProductionPolymers

Polymers

50 marksprintable — key stays hidden on paper

Level 3 Production Paper (Derivations & Explain-Out-Loud)

Time: 45 minutes Total Marks: 50

Instructions: Show all working. Where a derivation is requested, build the result from first definitions. Mechanisms must show the reactive intermediate and propagation step.


Q1. Molecular-weight statistics — derive from scratch. (12 marks)

A polymer sample contains only two species:

  • 4 mol of chains each of molar mass 10,000 g/mol10{,}000\ \text{g/mol}
  • 1 mol of chains each of molar mass 100,000 g/mol100{,}000\ \text{g/mol}

(a) Starting from the defining summation formulas, state Mˉn\bar{M}_n and Mˉw\bar{M}_w in terms of number of molecules NiN_i and mass MiM_i. (3)

(b) Compute Mˉn\bar{M}_n and Mˉw\bar{M}_w for this sample. (6)

(c) Compute the polydispersity index (PDI) and state what its value tells you physically. (3)


Q2. Free-radical polymerization mechanism — from memory. (10 marks)

For the polymerization of styrene initiated by benzoyl peroxide:

(a) Write the three chain stages (initiation, propagation, termination) with equations/structures, clearly showing the radical intermediate. (6)

(b) Explain out loud (in words) why polystyrene forms head-to-tail linkages preferentially, referencing radical stability and steric factors. (2)

(c) Distinguish termination by combination vs disproportionation and state which gives higher molecular weight. (2)


Q3. Condensation polymers — build the repeat unit. (10 marks)

(a) Draw the repeat unit of nylon-6,6 and name the two monomers, indicating which functional groups react and the small molecule eliminated. (4)

(b) Draw the repeat unit of terylene (PET) and name its two monomers. (3)

(c) Kevlar and Nomex are both aramids. Explain the structural difference between them and why Kevlar has superior tensile strength for body armour. (3)


Q4. Classification reasoning. (8 marks)

Classify EACH of the following on all THREE axes where applicable — (i) addition vs condensation, (ii) thermoplastic vs thermosetting, (iii) homopolymer vs copolymer — and justify in one line:

(a) Bakelite (2) (b) PVC (2) (c) Nylon-6,6 (2) (d) Melamine–formaldehyde resin (2)


Q5. Ziegler–Natta vs free-radical for polyethene. (6 marks)

(a) Name the catalyst system in Ziegler–Natta polymerization and state the type of mechanism (chain vs step; and the coordination step). (3)

(b) Explain why the polyethene made by Ziegler–Natta (HDPE) has higher density and tensile strength than that made by high-pressure free-radical routes (LDPE). (3)


Q6. Biodegradable & aerospace polymers — explain-out-loud. (4 marks)

(a) Name the two monomers of PHBV and state why it is biodegradable. (2)

(b) Give ONE aerospace polymer used for high-temperature service and ONE used as a composite matrix, with the property that justifies each choice. (2)

Answer keyMark scheme & solutions

Q1 (12 marks)

(a) Definitions (3) Mˉn=NiMiNi(1.5)\bar{M}_n = \frac{\sum N_i M_i}{\sum N_i} \quad (1.5) Mˉw=NiMi2NiMi(1.5)\bar{M}_w = \frac{\sum N_i M_i^2}{\sum N_i M_i} \quad (1.5) Why: Mˉn\bar M_n weights by number of molecules (mole fraction); Mˉw\bar M_w weights each molecule by its own mass, so heavy chains count more.

(b) Data: N1=4,M1=104N_1=4, M_1=10^4; N2=1,M2=105N_2=1, M_2=10^5. (6)

Ni=5\sum N_i = 5 NiMi=4(104)+1(105)=40,000+100,000=140,000\sum N_i M_i = 4(10^4)+1(10^5) = 40{,}000+100{,}000 = 140{,}000 Mˉn=140,0005=28,000 g/mol(3)\bar{M}_n = \frac{140{,}000}{5} = 28{,}000\ \text{g/mol} \quad (3)

NiMi2=4(104)2+1(105)2=4×108+1×1010=1.04×1010\sum N_i M_i^2 = 4(10^4)^2 + 1(10^5)^2 = 4\times10^8 + 1\times10^{10} = 1.04\times10^{10} Mˉw=1.04×1010140,000=74,285.7 g/mol(3)\bar{M}_w = \frac{1.04\times10^{10}}{140{,}000} = 74{,}285.7\ \text{g/mol} \quad (3)

(c) PDI (3) PDI=MˉwMˉn=74,285.728,000=2.653(2)\text{PDI} = \frac{\bar{M}_w}{\bar{M}_n} = \frac{74{,}285.7}{28{,}000} = 2.653 \quad (2) Since PDI > 1, the sample is polydisperse (broad distribution of chain lengths). A value ~2.65 indicates significant heterogeneity dominated by a few very heavy chains. (1)


Q2 (10 marks)

(a) (6 marks) — 2 marks each stage

Initiation: (C6H5COO)2Δ2C6H5COO2C6H5+2CO2(\text{C}_6\text{H}_5\text{COO})_2 \xrightarrow{\Delta} 2\,\text{C}_6\text{H}_5\text{COO}^\bullet \rightarrow 2\,\text{C}_6\text{H}_5^\bullet + 2\text{CO}_2; radical R\text{R}^\bullet adds to CH2=CHPh\text{CH}_2=\text{CHPh} giving R–CH2CHPh\text{R–CH}_2–\overset{\bullet}{\text{C}}\text{HPh}. Propagation: radical adds successively to monomer: ...CH2CHPh+CH2=CHPh...CH2CHPh–CH2CHPh\text{...CH}_2–\overset{\bullet}{\text{C}}\text{HPh} + \text{CH}_2=\text{CHPh} \rightarrow \text{...CH}_2–\text{CHPh–CH}_2–\overset{\bullet}{\text{C}}\text{HPh}. Termination: combination 2RR–R2\,\text{R}^\bullet \rightarrow \text{R–R} (or disproportionation).

(b) (2) The radical adds to the CH2\text{CH}_2 (less-substituted) end so the unpaired electron sits on the carbon bearing the phenyl group — a benzylic radical stabilized by resonance with the ring; adding at that carbon is also less sterically hindered. This regioselectivity repeats, giving head-to-tail chains.

(c) (2) Combination: two growing radicals couple into one dead chain → doubles molecular weight. Disproportionation: H-atom transfer gives two dead chains (one saturated, one with terminal C=C) of lower MW. Combination gives higher molecular weight. (1+1)


Q3 (10 marks)

(a) (4) Nylon-6,6 repeat unit: [NH(CH2)6NHCO(CH2)4CO]-[\text{NH}-(\text{CH}_2)_6-\text{NH}-\text{CO}-(\text{CH}_2)_4-\text{CO}]- Monomers: hexamethylenediamine (–NH₂ groups) + adipic acid (–COOH groups). Amide (peptide) linkage forms; water eliminated. (2 structure + 1 monomers + 1 by-product)

(b) (3) PET repeat unit: [OCH2CH2OCOC6H4CO]-[\text{O}-\text{CH}_2\text{CH}_2-\text{O}-\text{CO}-\text{C}_6\text{H}_4-\text{CO}]- Monomers: ethylene glycol + terephthalic acid (or dimethyl terephthalate). Ester linkage. (2 structure + 1 monomers)

(c) (3) Both are polyamides of a diamine + terephthalic/isophthalic acid. Kevlar = poly(para-phenylene terephthalamide): all-para linkages → rigid, rod-like linear chains that align and H-bond into ordered sheets → very high tensile strength. Nomex = meta-substituted → kinked chains, less ordered, so not as strong but excellent thermal/flame resistance. Kevlar's para geometry + interchain H-bonding gives the axial strength needed for body armour.


Q4 (8 marks)

(a) Bakelite: condensation; thermosetting (3-D cross-linked); copolymer (phenol + formaldehyde). (2) (b) PVC: addition; thermoplastic; homopolymer (vinyl chloride). (2) (c) Nylon-6,6: condensation; thermoplastic; copolymer (two different monomers). (2) (d) Melamine–formaldehyde: condensation; thermosetting (cross-linked network); copolymer. (2)

(1 mark each = axis classifications; 1 mark each = correct justification/monomer note)


Q5 (6 marks)

(a) (3) Catalyst: triethylaluminium Al(C2H5)3\text{Al(C}_2\text{H}_5)_3 + titanium tetrachloride TiCl4\text{TiCl}_4 (Ziegler–Natta). It is a chain-growth coordination polymerization: monomer coordinates to the vacant site on the transition-metal centre, then inserts into the metal–carbon bond (migratory insertion). (catalyst 1.5 + mechanism/coordination 1.5)

(b) (3) Ziegler–Natta gives linear, unbranched chains that pack closely and crystallize → higher density, higher tensile strength and melting point (HDPE). Free-radical high-pressure route causes chain-transfer/back-biting → branched chains that pack poorly → lower density, lower strength (LDPE).


Q6 (4 marks)

(a) (2) PHBV = copolymer of 3-hydroxybutanoic acid and 3-hydroxypentanoic (3-hydroxyvaleric) acid. Biodegradable because it contains hydrolysable ester linkages broken down by microorganisms/enzymes. (1 monomers + 1 reason)

(b) (2) High-temperature service: polyimides (retain properties at very high T) or Nomex (flame resistance). Composite matrix: epoxy resins (strong adhesion, cross-link into rigid CFRP/GFRP matrix). (1 + 1 with property)


[
  {"claim":"Mn = 28000 g/mol for the two-species sample","code":"N1,M1,N2,M2=4,10000,1,100000; Mn=(N1*M1+N2*M2)/(N1+N2); result = Mn==28000"},
  {"claim":"Mw = 74285.71... g/mol","code":"N1,M1,N2,M2=4,10000,1,100000; Mw=(N1*M1**2+N2*M2**2)/(N1*M1+N2*M2); result = simplify(Mw - Rational(1040000,14))==0"},
  {"claim":"PDI approx 2.653","code":"N1,M1,N2,M2=4,10000,1,100000; Mn=(N1*M1+N2*M2)/(N1+N2); Mw=(N1*M1**2+N2*M2**2)/(N1*M1+N2*M2); PDI=Mw/Mn; result = abs(float(PDI)-2.6531)<0.01"},
  {"claim":"Sum of NiMi = 140000","code":"result = (4*10000+1*100000)==140000"}
]